Management Science 461

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Management Science 461

• Lecture 4b P-median problems
• September 30, 2008

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Problem with coverage

• Coverage models are best for worst case
  problems
• We want to ensure good response for even the most
  remote demand node in the network
• Density does not drive the model, the lack of
  density does
• Central assumption if its close, its covered

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Problem with coverage

• Coverage model treats each demand node the same
  (max coverage the exception)
• A more appropriate measure is needed to find good
  average solutions
• This is what median problems are good for

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Example Network
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Problem Description

• Need to locate facilities and allocate customers
  to the facility so as to minimize total distance
  traveled
• Decision variables
• Locate at j or not binary value Xj
• Allocate customer i to facility j binary Yij

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Problem Description

• Cant allocate a customer to facility j if no
  facility located at j linking constraints
• Need to allocate each customer to a single
  facility
• Need to locate exactly P facilities

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Formulation
Cannot assign demands to an unopened facility
No. to locate Each demand assigned
once Integrality
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Median Solution for P1
100
250
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Locate at B for a total demand weighted distance
of 10,075
A
B
13
150
10
E
17
23
16
C
12
Demand
200
D
125
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General Formulation
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Extensions

• Facility costs
• Need to convert total travel to a cost to
  incorporate both in the same model
• Relax one customer, one facility
• Add a capacity constrainton facilities
• All these things make the problem harder than it
  already is

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Solving 1-Median

• Locating a single facility is straightforward
• Gravity model if location unrestricted and/or no
  network defined
• Enumeration of all candidate sites
• Demo in Excel

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Hakimi Proof
Each node has a weight (demand) of wA, wB, wC,
etc.
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Hakimi Proof
A
B
C
D
F
E
H
J
G
These nodes access the facility through F
These nodes access the facility through E
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Hakimi Proof
At this point, we can estimate the current cost
of moving demand from nodes E and F to the
facility location as (3 30) (7 40) 370
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Hakimi Proof
So having the facility at a3, b7 means total
travel is 370 km. But what happens when we
locate on node E (a0, b10)? What about a10,b0?
wB wC wF wJ (Say, 40 units)
wA wD wE wG wH (Say, 30 units)
F
E
a units
b units
Say a3, b7
Locate on E (0 30) (10 40) 400 ?
HIGHER Locate on F (10 30) (0 40) 300 ?
LOWER
Better solution if we move to the node with
higher weight (in this example, move to Node F)
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Hakimi Proof
What happens if both E and F have the same demand
of 35?
wB wC wF wJ (Say, 40 units)
wA wD wE wG wH (Say, 35 units)
F
E
a units
b units
Say a3, b7
a3, b7 335 735 350 Locate at either
node 1035 350 ? SAME
Property When demand at the two nodes is
different, can always get a better solution
moving to the node with higher weight. If equal
demand, all points along the arc (including both
nodes) are optimal. Thus, there is always an
optimal solution on the node.
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Complexity of P-median

• Original problem has n choose p solutions n! /
  (n-p)!p!
• For n10 and p3, 120 solutions
• For n100 and p15, 2.5E17 solutions
• At 1,000,000,000 solutions per second, how long
  would total enumeration take?
• non-polynomial problem heuristic solutions
  needed

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Solving the P-median Problem

• Greedy adding or Myopic algorithm
• Greedy algorithm with Substitution(Teitz and
  Bart, 1968, Operations Research)
• Neighborhood search(Maranzana, 1965, Operations
  Research Quarterly)
• Variable neighborhood search(Hansen and
  Mladenovic, 1997, Location Science)
• Lagrangian relaxation in BB

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The Teitz and Bart Heuristic

• Select a random solution
• Allocate the demand points to the selected
  facilities using shortest distances
• Compute the total cost of the current solution
• For each point A in the current solution
• For each point B, not in the current solution
• Consider replacing A with B
• Compute the total cost of the new solution (after
  replacement)
• If the new cost is less than the old cost,
  replace A with B, otherwise keep A in the solution

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Teitz-Bart worst case analysis

• For n100, p15
• At each iteration 15 points in the solution and
  85 points outside the solution
• Worst case Check 15851,275 solns per
  iteration
• Usually solved in lt 1000 iterations or faster